\begin{answer}
    For convenience we define
$$
\tilde w^{(i)}_j = 1\{\tilde z^{(i)} = j\}
$$
From the lecture note, we should easily derive the derivative from $\mu_j$ as
$$
\begin{aligned}
\nabla_{\mu_j}l &= \sum_{i=1}^m w^{(i)}_j \Sigma_j^{-1}(x^{(i)} - \mu_j) + \alpha \sum_{i=1}^{\tilde m}\tilde w^{(i)}_j \Sigma_j^{-1}(\tilde x^{(i)} - \mu_j)
\end{aligned}
$$
Setting this to zero, we get
$$
\mu_j = \frac{\sum_{i=1}^m w^{(i)}_j x^{(i)}+ \alpha \sum_{i=1}^{\tilde m}\tilde w^{(i)}_j\tilde x^{(i)}}{\sum_{i=1}^m w^{(i)}_j + \alpha \sum_{i=1}^{\tilde m}\tilde w^{(i)}_j}
$$
Similarly,
$$
\Sigma_j = \frac{\sum_{i=1}^m w^{(i)}_j (x^{(i)} - \mu_j)(x^{(i)} - \mu_j)^T+ \alpha \sum_{i=1}^{\tilde m}\tilde w^{(i)}_j (\tilde x^{(i)} - \mu_j)(\tilde x^{(i)} - \mu_j)^T}{\sum_{i=1}^m w^{(i)}_j + \alpha \sum_{i=1}^{\tilde m}\tilde w^{(i)}_j}
$$
For $\phi$, we need to maximize
$$
\sum_{i=1}^m\sum_{j=1}^kw^{(i)}_j \log \phi_j + \alpha\sum_{i=1}^{\tilde m}\sum_{j=1}^k\tilde w^{(i)}_j \log \phi_j
$$
subjective to
$$
\sum_{j=1}^k \phi_j = 1
$$
We construct the Lagrangian
$$
    \mathcal L(\phi, \beta) =\sum_{i=1}^m\sum_{j=1}^kw^{(i)}_j \log \phi_j + \alpha\sum_{i=1}^{\tilde m}\sum_{j=1}^k\tilde w^{(i)}_j \log \phi_j + \beta(1 - \sum_{j=1}^k \phi_j)
$$
And derivetive for
$$
    \frac{\partial\mathcal L}{\partial \phi_j} = \sum_{i=1}^m\frac{w^{(i)}_j}{\phi_j} + \alpha \sum_{i=1}^{\tilde m}\frac{\tilde w^{(i)}_j}{\phi_j} - \beta
$$
Set it to zero we get
$$
    \phi_j = \frac{\sum_{i=1}^mw^{(i)}_j + \alpha \sum_{i=1}^{\tilde m}{\tilde w}^{(i)}_j}{\beta}
$$
Considering the constraint $\sum_{j=1}^k \phi_j = 1$, we get $\beta = m + \alpha \tilde m$. So
$$
    \phi_j = \frac{\sum_{i=1}^mw^{(i)}_j + \alpha \sum_{i=1}^{\tilde m}\tilde w^{(i)}_j}{m + \alpha \tilde m}
$$



\end{answer}
